Electric Heating

A heating element converts electrical energy into heat by $I^2R$ loss, so its material must withstand high temperature for a long time. The desirable properties are:

1. High resistivity: a short length of wire gives the required resistance, so the element is compact.
2. High melting point: it can work at high temperature with a safety margin.
3. Low temperature coefficient of resistance: resistance (and so power) stays nearly constant from cold to hot.
4. Free from oxidation at high temperature: oxidation reduces cross-section and causes early burnout.
5. High mechanical strength and ductility: it can be drawn into wire or strip and wound into coils, and it does not break when heated and cooled repeatedly.
6. Low cost and easy availability.

Common materials: nichrome (80% Ni, 20% Cr, up to about 1150°C), Kanthal (Fe-Cr-Al, up to about 1300°C), silicon carbide (up to about 1450°C), molybdenum and tungsten (in vacuum or inert gas, above 1500°C).

In a low-frequency (core-type) induction furnace the charge forms the short-circuited secondary, so the secondary current is set by the secondary resistance $R$ and reactance $X$.

Given: $V_2 = 20$ V, $P = 500$ kW at pf 0.5 (hearth full). Half-full: $R' = 2R$, $X$ unchanged.

Secondary impedance when hearth is full

$$\begin{aligned} S &= \frac{P}{\cos\phi} = \frac{500}{0.5} = 1000\ \text{kVA} \\ I_2 &= \frac{S}{V_2} = \frac{1000 \times 10^3}{20} = 50\,000\ \text{A} \\ Z &= \frac{V_2}{I_2} = \frac{20}{50\,000} = 4 \times 10^{-4}\ \Omega \\ R &= Z\cos\phi = 2 \times 10^{-4}\ \Omega \\ X &= Z\sin\phi = 4\times10^{-4} \times 0.866 = 3.464 \times 10^{-4}\ \Omega \end{aligned}$$

Hearth half full

$$\begin{aligned} R' &= 2R = 4 \times 10^{-4}\ \Omega, \quad X = 3.464 \times 10^{-4}\ \Omega \\ Z' &= \sqrt{(4)^2 + (3.464)^2} \times 10^{-4} = 5.2915 \times 10^{-4}\ \Omega \\ I_2' &= \frac{20}{5.2915 \times 10^{-4}} = 37\,796\ \text{A} \\ P' &= I_2'^2 R' = 37\,796^2 \times 4 \times 10^{-4} = 571.4\ \text{kW} \\ \cos\phi' &= \frac{R'}{Z'} = \frac{4}{5.2915} = 0.756\ \text{lagging} \end{aligned}$$

Answer: With the hearth half full, power absorbed $\approx 571.4$ kW at a power factor of $0.756$ lagging.

Each phase of the arc furnace is the arc (a pure resistance) in series with the transformer resistance and reactance.

Given (per phase): $I = 4000$ A, $V_{arc} = 60$ V, $R_t = 0.0025\ \Omega$, $X_t = 0.0050\ \Omega$.

i) Power factor and kW drawn

$$\begin{aligned} R_{arc} &= \frac{V_{arc}}{I} = \frac{60}{4000} = 0.015\ \Omega \\ R_{total} &= 0.015 + 0.0025 = 0.0175\ \Omega \\ Z &= \sqrt{0.0175^2 + 0.005^2} = 0.01820\ \Omega \\ \cos\phi &= \frac{R_{total}}{Z} = \frac{0.0175}{0.01820} = 0.9615\ \text{lagging} \\ P &= 3I^2R_{total} = 3 \times 4000^2 \times 0.0175 = 840\ \text{kW} \end{aligned}$$

(Arc power alone $= 3 \times 60 \times 4000 = 720$ kW; the other 120 kW is transformer copper loss.)

ii) Time to melt 2.5 tonnes of steel

Heat required:

$$\begin{aligned} H &= m\left[c(\theta_m - \theta_i) + L\right] \\ &= 2500\left[0.5(1370 - 15) + 37.2\right] \\ &= 2500 \times 714.7 = 1\,786\,750\ \text{kJ} \end{aligned}$$

Useful power with 70% overall efficiency:

$$P_{useful} = 0.7 \times 840 = 588\ \text{kW}$$

Time:

$$t = \frac{1\,786\,750}{588} = 3038.7\ \text{s} = 50.6\ \text{min}$$

Answer: pf $= 0.9615$ lagging, power drawn $= 840$ kW; time to melt 2.5 t of steel $\approx 3039$ s $\approx 50.6$ minutes.

The wire must have the resistance that gives 4.5 kW at 200 V, and enough surface area to radiate that power at the given temperatures.

Given: $P = 4500$ W, $V = 200$ V, $T_1 = 1000 + 273 = 1273$ K, $T_2 = 500 + 273 = 773$ K, $\rho = 42.5 \times 10^{-6}\ \Omega$m (as given), $k = 1.0$, $e = 0.9$.

Heat dissipated per unit surface area (Stefan's law)

$$\begin{aligned} H &= 5.72\,e\,k\left[\left(\frac{T_1}{100}\right)^4 - \left(\frac{T_2}{100}\right)^4\right]\ \text{W/m}^2 \\ &= 5.72 \times 0.9 \times 1.0\,[12.73^4 - 7.73^4] \\ &= 5.148 \times (26\,261 - 3\,570) = 1.168 \times 10^5\ \text{W/m}^2 \end{aligned}$$

Resistance condition

$$R = \frac{V^2}{P} = \frac{200^2}{4500} = 8.889\ \Omega, \qquad R = \frac{4\rho l}{\pi d^2}$$ $$\frac{l}{d^2} = \frac{\pi R}{4\rho} = \frac{\pi \times 8.889}{4 \times 42.5\times10^{-6}} = 1.6427 \times 10^5 \quad (1)$$

Heat-dissipation condition

$$P = \pi d l H \Rightarrow d\,l = \frac{4500}{\pi \times 1.168 \times 10^5} = 0.012262 \quad (2)$$

Solving

Dividing (2) by (1): $d^3 = \dfrac{0.012262}{1.6427 \times 10^5} = 7.465 \times 10^{-8}$

$$\begin{aligned} d &= 4.21 \times 10^{-3}\ \text{m} = 4.21\ \text{mm} \\ l &= \frac{0.012262}{4.21 \times 10^{-3}} = 2.91\ \text{m} \end{aligned}$$

Answer: Diameter $\approx 4.21$ mm, length $\approx 2.91$ m.

Note: the given resistivity is used as printed; real nichrome is about $1.1\ \mu\Omega$m, which would give a thinner, longer wire.

The wire must give 5.5 kW at 220 V (resistance condition) and must radiate this power at 1100°C (surface condition).

Given: $P = 5500$ W, $V = 220$ V, $T_1 = 1100 + 273 = 1373$ K, $T_2 = 500 + 273 = 773$ K, $\rho = 42.5 \times 10^{-6}\ \Omega$m (as given), $k = 1.0$, $e = 0.9$.

Heat dissipated per m² of wire surface

$$\begin{aligned} H &= 5.72\,e\,k\left[\left(\frac{T_1}{100}\right)^4 - \left(\frac{T_2}{100}\right)^4\right] \\ &= 5.72 \times 0.9 \times 1.0 \times (13.73^4 - 7.73^4) \\ &= 5.148 \times (35\,537 - 3\,570) = 1.6456 \times 10^5\ \text{W/m}^2 \end{aligned}$$

Resistance condition

$$R = \frac{V^2}{P} = \frac{220^2}{5500} = 8.8\ \Omega$$ $$R = \frac{\rho l}{\pi d^2/4} \Rightarrow \frac{l}{d^2} = \frac{\pi R}{4\rho} = \frac{\pi \times 8.8}{4 \times 42.5 \times 10^{-6}} = 1.6262 \times 10^5 \quad (1)$$

Surface (heat) condition

$$P = (\pi d l)H \Rightarrow d\,l = \frac{5500}{\pi \times 1.6456\times10^5} = 0.010638 \quad (2)$$

Solving (1) and (2)

$$\begin{aligned} d^3 &= \frac{0.010638}{1.6262 \times 10^5} = 6.542 \times 10^{-8} \\ d &= 4.03 \times 10^{-3}\ \text{m} = 4.03\ \text{mm} \\ l &= \frac{0.010638}{4.03 \times 10^{-3}} = 2.64\ \text{m} \end{aligned}$$

Answer: Wire diameter $\approx 4.03$ mm and length $\approx 2.64$ m.

Note: the resistivity is used as given; real nichrome is about $1.1\ \mu\Omega$m.

For a thin strip, the radiating surface is both faces, $2wl$ (edges neglected), and the cross-section is $wt$.

Given: $P = 40$ kW, 3-phase, 400 V, star connected; $t = 0.0025$ cm $= 2.5 \times 10^{-5}$ m; $T_1 = 1100 + 273 = 1373$ K; $T_2 = 700 + 273 = 973$ K; $k = 0.6$; $e = 0.9$; $\rho = 1.03 \times 10^{-6}\ \Omega$m.

Per-phase values

$$\begin{aligned} P_{ph} &= \frac{40\,000}{3} = 13\,333\ \text{W}, \quad V_{ph} = \frac{400}{\sqrt3} = 230.94\ \text{V} \\ R &= \frac{V_{ph}^2}{P_{ph}} = \frac{230.94^2}{13\,333} = 4.0\ \Omega \end{aligned}$$

Heat dissipated per m² (Stefan's law)

$$\begin{aligned} H &= 5.72\,e\,k\left[\left(\frac{T_1}{100}\right)^4 - \left(\frac{T_2}{100}\right)^4\right] \\ &= 5.72 \times 0.9 \times 0.6 \times (13.73^4 - 9.73^4) \\ &= 3.0888 \times (35\,537 - 8\,963) = 82\,082\ \text{W/m}^2 \end{aligned}$$

Equations for the strip

$$\begin{aligned} R = \frac{\rho l}{w t} &\Rightarrow \frac{l}{w} = \frac{R t}{\rho} = \frac{4 \times 2.5\times10^{-5}}{1.03\times10^{-6}} = 97.09 \quad (1) \\ P_{ph} = 2wlH &\Rightarrow wl = \frac{13\,333}{2 \times 82\,082} = 0.08122\ \text{m}^2 \quad (2) \end{aligned}$$

Solving

$$\begin{aligned} w^2 &= \frac{0.08122}{97.09} = 8.365 \times 10^{-4} \\ w &= 0.02892\ \text{m} = 28.9\ \text{mm} \\ l &= 97.09 \times 0.02892 = 2.81\ \text{m (per phase)} \end{aligned}$$

Answer: Strip width $\approx 28.9$ mm (about 2.9 cm), with length $\approx 2.81$ m per phase.

Characteristics of heating elements

A heating element produces heat by $I^2R$ loss, so its material should have:

1. High resistivity — a short, compact element gives the needed resistance. Example: nichrome ($\approx 1.1\ \mu\Omega$m) is about 60 times copper.
2. High melting point — to work at high temperature. Example: tungsten (3400°C) for very high-temperature furnaces.
3. Low temperature coefficient of resistance — power stays steady as the element heats. Example: nichrome changes very little from cold to hot.
4. Resistance to oxidation — no scaling at working temperature. Example: nichrome forms a protective chromium-oxide layer; molybdenum must be used in inert gas.
5. Mechanical strength and ductility — can be drawn into wire or strip and survives thermal cycling.
6. Low cost.

Common elements: nichrome (up to 1150°C), Kanthal (up to 1300°C), silicon carbide rods (up to 1450°C).

Numerical: 25 kW, 240 V nichrome heater

Given: $T_1 = 1400 + 273 = 1673$ K, $T_2 = 500 + 273 = 773$ K, $\rho = 42.5 \times 10^{-6}\ \Omega$m (as given), $k = 0.9$, $e = 0.85$.

Heat dissipated per m²:

$$\begin{aligned} H &= 5.72\,e\,k\left[\left(\frac{T_1}{100}\right)^4 - \left(\frac{T_2}{100}\right)^4\right] \\ &= 5.72 \times 0.85 \times 0.9 \times (16.73^4 - 7.73^4) \\ &= 4.3758 \times (78\,340 - 3\,570) = 3.272 \times 10^5\ \text{W/m}^2 \end{aligned}$$

Resistance condition:

$$R = \frac{240^2}{25\,000} = 2.304\ \Omega, \qquad \frac{l}{d^2} = \frac{\pi R}{4\rho} = \frac{\pi \times 2.304}{4 \times 42.5\times10^{-6}} = 42\,578 \quad (1)$$

Surface condition:

$$P = \pi d l H \Rightarrow d\,l = \frac{25\,000}{\pi \times 3.272\times10^5} = 0.024322 \quad (2)$$

Solving:

$$\begin{aligned} d^3 &= \frac{0.024322}{42\,578} = 5.712 \times 10^{-7} \\ d &= 8.30 \times 10^{-3}\ \text{m} = 8.30\ \text{mm} \\ l &= \frac{0.024322}{8.30 \times 10^{-3}} = 2.93\ \text{m} \end{aligned}$$

Answer: Diameter $\approx 8.30$ mm, length $\approx 2.93$ m (using the resistivity as printed).

The board's thickness is missing from the question; a thickness of 2 cm is assumed (board 30 cm × 25 cm × 2 cm). The current does not depend on this assumption (see the note at the end).

Given: $A = 0.30 \times 0.25 = 0.075$ m², $d = 0.02$ m, $\Delta\theta = 170 - 20 = 150$°C, $t = 10$ min $= 600$ s, $f = 30$ MHz, $c = 0.35$ cal/g°C, density $= 0.55$ g/cc, $\varepsilon_r = 5$, pf $= 0.05$, losses 15% (efficiency 85%).

Heat and power required

$$\begin{aligned} \text{Volume} &= 30 \times 25 \times 2 = 1500\ \text{cm}^3, \quad m = 0.55 \times 1500 = 825\ \text{g} \\ H &= m c \Delta\theta = 825 \times (0.35 \times 4.186) \times 150 = 181\,306\ \text{J} \\ P_{out} &= \frac{181\,306}{600} = 302.2\ \text{W} \\ P_{in} &= \frac{302.2}{0.85} = 355.5\ \text{W} \end{aligned}$$

Capacitance of the work piece

$$C = \frac{\varepsilon_0 \varepsilon_r A}{d} = \frac{8.854\times10^{-12} \times 5 \times 0.075}{0.02} = 166.0\ \text{pF}$$

Voltage across the work

Dielectric loss: $P = V^2 \omega C \cos\phi$

$$\begin{aligned} V &= \sqrt{\frac{P}{2\pi f C \cos\phi}} = \sqrt{\frac{355.5}{2\pi \times 30\times10^6 \times 166.0\times10^{-12} \times 0.05}} \\ &= \sqrt{\frac{355.5}{1.5646 \times 10^{-3}}} = 476.7\ \text{V} \end{aligned}$$

Current

$$I = \frac{P}{V\cos\phi} = \frac{355.5}{476.7 \times 0.05} = 14.92\ \text{A}$$

Answer (for 2 cm thickness): Power input $\approx 355.5$ W, voltage $\approx 477$ V, current $\approx 14.9$ A.

Note: $V$ is proportional to the thickness (for 1 cm, $V \approx 238$ V; for 2.5 cm, $V \approx 596$ V), but the current stays 14.9 A because both the power and the capacitance scale with thickness.

In a low-frequency core-type induction furnace the charge forms a single-turn secondary, so its current depends on the secondary resistance $R$ and reactance $X$.

Given: $V_2 = 10$ V, $P = 500$ kW at pf 0.5 (hearth full); half-full: $R' = R/2$, $X$ unchanged.

Hearth full

$$\begin{aligned} S &= \frac{500}{0.5} = 1000\ \text{kVA}, \quad I_2 = \frac{1000\times10^3}{10} = 100\,000\ \text{A} \\ Z &= \frac{10}{100\,000} = 1.0 \times 10^{-4}\ \Omega \\ R &= Z\cos\phi = 0.5\times10^{-4}\ \Omega, \quad X = Z\sin\phi = 0.866 \times 10^{-4}\ \Omega \end{aligned}$$

Hearth half full

$$\begin{aligned} R' &= 0.25 \times 10^{-4}\ \Omega \\ Z' &= \sqrt{0.25^2 + 0.866^2} \times 10^{-4} = 0.9014 \times 10^{-4}\ \Omega \\ I_2' &= \frac{10}{0.9014 \times 10^{-4}} = 110\,940\ \text{A} \\ P' &= I_2'^2 R' = 110\,940^2 \times 0.25 \times 10^{-4} = 307.7\ \text{kW} \\ \cos\phi' &= \frac{R'}{Z'} = \frac{0.25}{0.9014} = 0.277\ \text{lagging} \end{aligned}$$

Answer: Power absorbed $\approx 307.7$ kW at power factor $\approx 0.277$ lagging. Halving the resistance raises the current but lowers both the power and the power factor, because the circuit becomes mainly reactive.

Induction heating heats a conducting charge by eddy currents (and hysteresis loss in magnetic materials) induced in it by an alternating magnetic field; the charge acts as the short-circuited secondary of a transformer.

Advantages of induction heating

1. Heat is produced inside the charge itself, so heating is fast and efficient.
2. No contact between coil and charge, so no contamination; pure metals and alloys can be melted.
3. Accurate temperature control and uniform heating; electromagnetic stirring mixes the melt well.
4. Selective heating of a chosen depth or area by choosing frequency (skin effect), e.g. surface hardening.
5. Clean, no smoke, no fuel storage, quiet and safe working conditions.
6. Quick start, low heat loss to surroundings and low running cost for large outputs.

Core-type (low-frequency) induction furnace

        +------------------+
        |   laminated core |
   +----+----+        +----+----+
   | primary |        | molten  |
   |  coil   |  core  | metal   |
   | (50 Hz) |        | ring =  |
   +----+----+        | secondary
        |             +----+----+
        +------------------+

• A primary coil on a laminated iron core is fed at supply (or low) frequency.
• The charge, in an annular or V-shaped channel around the core, forms a single-turn short-circuited secondary.
• Large current flows in the charge, heating it by $I^2R$.
• Problems: poor pf (high leakage reactance), pinch effect at high current density, and the furnace must be started with molten metal in the ring. The Ajax-Wyatt (vertical core) furnace overcomes many of these.

Coreless (high-frequency) induction furnace

        water-cooled copper coil
        o  +--------------+  o
        o  |   crucible   |  o
        o  |  [ charge ]  |  o   <- HF supply
        o  |              |  o      (0.5-10 kHz)
        o  +--------------+  o

• A crucible containing the charge is surrounded by a water-cooled copper coil fed from a high-frequency source (500 Hz–10 kHz).
• There is no iron core; the high frequency produces strong eddy currents in the charge.
• Capacitors are connected across the coil to correct the very low power factor.
• Melting is fast, with good stirring and accurate control.

Fields of application

• Melting of steel, copper, brass, aluminium and special alloys in foundries.
• Surface hardening of gears, shafts and crankshafts.
• Brazing, soldering, annealing and forging billets.
• Zone refining of semiconductors, sterilising, and induction cooktops.

Electric heating converts electrical energy into heat (by resistance, induction, arc or dielectric loss) for industrial, commercial and domestic uses.

Advantages of electric heating

1. Clean: no smoke, ash or flue gases; no need for chimneys or fuel storage.
2. Accurate temperature control by switches, thermostats and electronic controllers.
3. High efficiency: heat is produced directly in or near the charge; little is lost.
4. Uniform heating and heating of any desired part or depth (induction, dielectric).
5. Very high temperatures possible (arc furnace about 3000°C).
6. Quick starting, compact equipment and low labour cost.
7. Safe to operate; can be automated.
8. Heat can be produced in non-conducting materials (dielectric heating) and inside metals (induction heating).

Building design considerations for electric heating

The aim is to keep the heating load (and electricity bill) low while maintaining comfort.

1. Thermal insulation: walls, roof and floor should be insulated (low U-value) to reduce conduction loss; insulated roofs are most important since warm air rises.
2. Windows and glazing: double glazing, proper window area and orientation; south-facing windows in Nepal for solar gain in winter.
3. Air tightness and ventilation: seal gaps to reduce infiltration, but give controlled ventilation for fresh air; heat-recovery ventilation where possible.
4. Thermal mass: heavy materials (brick, stone, concrete) store heat and smooth temperature swings; useful with off-peak storage heaters.
5. Building orientation and shape: compact shape reduces surface area per volume; orientation to use sun and avoid cold wind.
6. Heat-loss calculation: room-by-room heat loss (conduction + ventilation) to size heaters correctly.
7. Choice and location of heaters: convectors, radiant heaters, under-floor or storage heaters placed under windows or near outer walls.
8. Controls and tariff: thermostats, timers and zoning; use off-peak/time-of-day tariff for storage heating.
9. Wiring capacity: circuits and protection rated for heater load; separate circuits for heavy heaters.

The initial temperature of the board is not given; 20°C is assumed. Calculations are done for efficiency 85% (the lower limit, which gives the largest power, used for sizing), with 90% shown for comparison.

Given: board $0.3 \times 0.15 \times 0.025$ m, final temperature 160°C, $t = 600$ s, $f = 30$ MHz, $c = 1465$ J/kg°C, density $= 575$ kg/m³, $\varepsilon_r = 5$, pf $= 0.05$.

Mass and heat required

$$\begin{aligned} m &= 575 \times (0.3 \times 0.15 \times 0.025) = 0.6469\ \text{kg} \\ H &= m c \Delta\theta = 0.6469 \times 1465 \times (160 - 20) = 132\,674\ \text{J} \\ P_{out} &= \frac{132\,674}{600} = 221.1\ \text{W} \end{aligned}$$

Power required

$$P_{in} = \frac{221.1}{0.85} = 260.1\ \text{W} \quad (\text{at } 90\%: 245.7\ \text{W})$$

Capacitance of the board

$$C = \frac{\varepsilon_0 \varepsilon_r A}{d} = \frac{8.854\times10^{-12} \times 5 \times (0.3\times0.15)}{0.025} = 79.69\ \text{pF}$$

Voltage across the work

Using $P = V^2\,2\pi f C \cos\phi$:

$$\begin{aligned} 2\pi f C \cos\phi &= 2\pi \times 30\times10^6 \times 79.69\times10^{-12} \times 0.05 = 7.510\times10^{-4} \\ V &= \sqrt{\frac{260.1}{7.510\times10^{-4}}} = 588.5\ \text{V} \end{aligned}$$

Current through the work

$$I = \frac{P}{V\cos\phi} = \frac{260.1}{588.5 \times 0.05} = 8.84\ \text{A}$$

Answer (85%): Power $\approx 260$ W, voltage $\approx 589$ V, current $\approx 8.84$ A.

Electric space heating of a building should keep rooms comfortable with the minimum heating load and energy cost. The heater rating depends on the heat lost from the building, so the design must reduce losses and use heat well.

Factors to be considered

1. Heat loss by conduction (fabric loss): loss through walls, roof, floor, windows and doors, $Q = U A \Delta\theta$. Use materials with low thermal transmittance (U-value) and add insulation, especially in the roof.
2. Ventilation and infiltration loss: heat carried away by air changes, $Q = 0.33\,n\,V\,\Delta\theta$ W (n air changes per hour, V room volume in m³). Seal cracks and use weather-stripping but keep enough fresh air.
3. Windows and glazing: glass loses heat fast; use double glazing, curtains, and suitable window area.
4. Orientation and solar gain: face main rooms and windows to the south (in Nepal) to gain winter sun; shield from cold winds.
5. Thermal capacity (mass) of the building: heavy walls store heat; good for storage heaters charged at off-peak hours and for steady temperature.
6. Design temperatures: required indoor temperature (about 18–22°C for living rooms) and lowest outdoor temperature of the site decide $\Delta\theta$ and heater size.
7. Occupancy and internal heat gains: heat from people, lights and appliances reduces the heating need.
8. Type and placement of heaters: convection, radiant, under-floor, storage or heat-pump systems; place heaters near cold surfaces (under windows) for even temperature.
9. Controls: thermostats, timers and zone control avoid overheating and waste.
10. Tariff and supply: time-of-day tariff favours off-peak storage heating; wiring and protection must suit the heating load.

   heat in (heater)  -->  room at 20 C
        |                    |
        |    losses: walls, roof, floor,
        |    windows, air leakage
        v                    v
   heater kW = fabric loss + ventilation loss
               - internal gains

Good insulation, air tightness and correct heater sizing together cut both the connected load and the running cost.

Interpretation: "resistance/reactance of temperature" means the transformer referred to the secondary, "overall efficiency of steel" means overall efficiency of the furnace, and specific heat is 0.5 kJ/kg°C. Values are per phase.

Given: $m = 4300$ kg/h, $c = 0.5$ kJ/kg°C, $L = 37.2$ kJ/kg, $\theta_m = 1370$°C, $\theta_i = 19.1$°C, $\eta = 0.5$, $I = 5700$ A, $R_t = 0.008\ \Omega$, $X_t = 0.014\ \Omega$.

(i) Average kW input

$$\begin{aligned} H &= m[c(\theta_m - \theta_i) + L] = 4300[0.5(1370 - 19.1) + 37.2] \\ &= 4300 \times 712.65 = 3\,064\,395\ \text{kJ/h} \\ P_{out} &= \frac{3\,064\,395}{3600} = 851.2\ \text{kW} \\ P_{in} &= \frac{851.2}{0.5} = 1702.4\ \text{kW} \end{aligned}$$

(ii) and (iii) Arc resistance and arc voltage

Per phase, all input power is dissipated in $R_t + R_{arc}$:

$$\begin{aligned} R_t + R_{arc} &= \frac{P_{in}}{3I^2} = \frac{1\,702\,442}{3 \times 5700^2} = 0.017466\ \Omega \\ R_{arc} &= 0.017466 - 0.008 = 0.009466\ \Omega \\ V_{arc} &= I R_{arc} = 5700 \times 0.009466 = 53.96\ \text{V} \end{aligned}$$

(iv) Power factor

$$\begin{aligned} Z &= \sqrt{0.017466^2 + 0.014^2} = 0.022385\ \Omega \\ \cos\phi &= \frac{0.017466}{0.022385} = 0.780\ \text{lagging} \end{aligned}$$

(v) Average kVA input

$$S = \frac{P_{in}}{\cos\phi} = \frac{1702.4}{0.7803} = 2181.8\ \text{kVA}$$

Answer: (i) 1702.4 kW, (ii) arc voltage $\approx 53.96$ V, (iii) arc resistance $\approx 0.00947\ \Omega$, (iv) pf $\approx 0.780$ lagging, (v) $\approx 2182$ kVA.

Assumptions: the strip radiates from both faces (edges neglected, surface $= 2wl$); the elements are star connected so each phase gets $V_L/\sqrt3$; the charge absorbs heat as a black body at 700°C; values are used as given ($t = 0.025$ cm, $\rho = 10.3\times10^{-6}\ \Omega$m).

Given: $P = 30$ kW, 400 V, 3-phase; $T_1 = 1373$ K, $T_2 = 973$ K; $k = 0.6$, $e = 0.9$; $t = 2.5\times10^{-4}$ m.

Per-phase values

$$\begin{aligned} P_{ph} &= \frac{30\,000}{3} = 10\,000\ \text{W}, \quad V_{ph} = \frac{400}{\sqrt3} = 230.94\ \text{V} \\ R &= \frac{230.94^2}{10\,000} = 5.333\ \Omega \end{aligned}$$

Heat dissipated per m²

$$\begin{aligned} H &= 5.72\,e\,k\left[\left(\frac{T_1}{100}\right)^4 - \left(\frac{T_2}{100}\right)^4\right] \\ &= 5.72 \times 0.9 \times 0.6 \times (13.73^4 - 9.73^4) \\ &= 3.0888 \times 26\,574 = 82\,082\ \text{W/m}^2 \end{aligned}$$

Strip equations

$$\begin{aligned} R = \frac{\rho l}{w t} &\Rightarrow \frac{l}{w} = \frac{R t}{\rho} = \frac{5.333 \times 2.5\times10^{-4}}{10.3\times10^{-6}} = 129.45 \quad (1)\\ P_{ph} = 2 w l H &\Rightarrow w l = \frac{10\,000}{2 \times 82\,082} = 0.060915\ \text{m}^2 \quad (2) \end{aligned}$$

Solving

$$\begin{aligned} w^2 &= \frac{0.060915}{129.45} = 4.706\times10^{-4} \\ w &= 0.02169\ \text{m} = 21.7\ \text{mm} \\ l &= 129.45 \times 0.02169 = 2.81\ \text{m per phase} \end{aligned}$$

Answer: Suitable strip width $\approx 21.7$ mm (about 2.2 cm), length $\approx 2.81$ m per phase.

All three heat the material without direct contact, but they differ in how and where the heat is produced.

• Dielectric heating: a non-conducting material placed between two electrodes at high frequency (10–50 MHz) is heated by dielectric loss (molecular polarisation). Heat is produced uniformly throughout the thickness.
• Infrared heating: tungsten-filament lamps (about 2300°C) with reflectors radiate infrared energy onto the surface; heat is absorbed at the surface and conducted inward.
• Microwave heating: a magnetron produces microwaves (usually 2.45 GHz) that make polar (water) molecules rotate rapidly; heat is produced inside the material in a metal cavity.

 Dielectric        Infrared          Microwave
 +--------+       \ | /             +----------+
 |electrode|       lamp             | magnetron|
 | [work] |       (IR)              |  ~~~>    |
 |electrode|      v v v             |  [food]  |
 +--------+      [surface]          +----------+

Electric heating converts electrical energy into heat. The main methods are classified as power-frequency and high-frequency heating.

              Electric heating
        +-----------+-----------+
   Power frequency          High frequency
   |-- Resistance           |-- Induction
   |     direct/indirect    |     core / coreless
   |-- Arc                  |-- Dielectric
         direct/indirect    |-- Microwave
                            |-- Infrared

1. Resistance heating

Heat is produced by $I^2R$ loss.

• Direct: current passes through the charge itself (e.g. salt-bath furnace, electrode boiler, resistance welding).
• Indirect: current passes through a heating element (nichrome, Kanthal) and heat reaches the charge by radiation and convection. Examples: room heaters, electric irons, kettles, water heaters, cooking ranges, ovens (domestic) and resistance ovens for annealing and drying (industrial).

2. Arc heating

An electric arc between electrodes produces about 3000–3500°C.

• Direct arc: arc between electrodes and charge; current flows through the charge (steel-making arc furnace).
• Indirect arc: arc between two electrodes above the charge; heat reaches it by radiation (melting of brass, copper).
• Also used in arc welding.

3. Induction heating

Eddy currents (and hysteresis loss) are induced in a conducting charge by an alternating field.

• Core type (low frequency): Ajax-Wyatt furnace for melting brass and non-ferrous metals.
• Coreless (high frequency): steel and alloy melting, surface hardening, brazing; induction cooktops in homes.

4. Dielectric heating

Non-conducting materials between electrodes at 10–50 MHz are heated by dielectric loss, uniformly through the volume. Used for gluing plywood, welding plastics, drying textiles and food processing.

5. Infrared heating

Infrared lamps radiate heat on a surface. Used for drying paints and varnishes, and in some domestic heaters.

6. Microwave heating

A magnetron (2.45 GHz) heats water-containing material from inside. Used in microwave ovens at home and industrial food drying.